Does $$\frac{1}{\Big|e^{\frac{2\pi a\sqrt{-1}}q} - 1\Big|} < q^{1/2}*(\log q)^2$$ always hold at any $0 Is there a proof? Can this be tightened to $$\frac{1}{\Big|e^{\frac{2\pi a\sqrt{-1}}q} - 1\Big|} < cq^{1/2}*\log q$$ at some fixed $c>0$?
A simple upper bound.
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inequality
1 Answers
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No. Take $a=1$, and let $q\to\infty$.
$$\lvert e^{\frac{2i\pi}{q}} - 1\rvert = \left\lvert \frac{2i\pi}{q} + o(\left(\frac{1}{q}\right) \right\rvert \operatorname*{\sim}_{q\to\infty} \frac{2\pi}{q}$$ so the LHS is asymptotically equivalent to $\frac{q}{2\pi}$. Since the RHS is $ \sqrt{q}\log^2 q = o(q) $, the inequality cannot hold when $q$ grows large.
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0would you know similar problem http://math.stackexchange.com/questions/2108106/a-simple-upper-bound-2? – 2017-01-22
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0@ao. Why did you delete that other question?( I was in the process of typing an answer.) – 2017-01-22
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0I thought there was a mistake. Sorry brought back up. – 2017-01-22